Lecture 4 Linear Programming Models: Standard Form

1 Lecture 4 Linear Programming Models: Standard FormAugust 31, 2009 Lecture 4 Outline: Standard form LP Transforming the LP problem to Standard form Basic solutions of Standard LP problemOperations Research Methods1 Lecture 4 Why Standard Form? The simplex method had proven to be the most efficient (practical)solver of LP problems The implementation of simplex method requires the LP problem instandard formOperations Research Methods2 Lecture 4 What is the Standard Form It is the LP model with the specific form of theconstraints:max (or min)z=c1x1+c2x2+ cnxnsubject toa11x1+a12x2+ +a1nxn=b1a21x1+a22x2+ +a2nxn= +am2x2+ +amnxn=bmx1 0, x2 0, .. , xn 0 mequalities andnnonnegativity constraints withm nOperations Research Methods3 Lecture 4 Bringing an LP to its Standard Form The inequality Introduce asurplus variable The inequality Introduce aslack variableNOTE: Thecost of surplus and slack variables is zero Unrestricted variable in sign:Replace it with adifference of two new variables All new variables have to benonnegativeOperations Research Methods4 Lecture 4 Exampleminimizez= 3x1+ 8x2+ 4x3subject tox1+x2 82x1 3x2 0x2 9x1, x2 0 Operations Research Methods5 Lecture 4 Its Standard form:minimizez= 3x1+ 8x2+ 4x7 4x8subject tox1+x2 x4= 82x1 3x2+x5= 0x2 x6= 9x1, x2, x4, x5, x6, x7, x8 0x3is substituted out of the problemWe have.

2 M= 3andn= 7 Operations Research Methods6 Lecture 4 What are Basic Solutions? Illustration on Reddy Mikks Constraint setBasic solutions are the corner pointsOperations Research Methods7 Lecture 4 What are the basic solutions? For a problem in thestandard form a basic solution is a point x= ( x1, .. , xn)that has at leastn mcoordinates equal to 0,and satisfies all the equality constraintsof the problema11 x1+a12 x2+ +a1n xn=b1a21 x1+a22 x2+ +a2n xn= x1+am2 x2+ +amn xn=bm If the point xhas all components nonnegative, , xi 0for alli, then xis abasic feasible solution Otherwise, ( , if xj<0for some indexj), xisbasic infeasiblesolutionOperations Research Methods8 Lecture 4 How to find the basic solutions algebraically If the problem is not in Standard form, bring it to the Standard form Basic solutions are determined from the Standard form as follows: Selectn mout ofnnonnegative inequalities (coordinate indices)i,xi 0,i= 1.

3 , mand set them to zeroxj= 0for a total ofn mindicesj(nonbasic variables) Substitute these zero values in the equalities:we havemunknown variables anmequalities Solve thism msystem of equations: we obtain values for the remainingmvariables (basic variables) If thesem(basic) variables are nonnegative, we have a basic feasiblesolution; otherwise, a basic infeasible solutionOperations Research Methods9 Lecture 4 Example Reddy Mikks ProblemOriginal LP formulationmaximizez= 5x1+ 4x2subject to6x1+ 4x2 24x1+ 2x2 6x1, x2 0 Standard LP formmaximizez= 5x1+ 4x2subject to6x1+ 4x2+x3= 24x1+ 2x2+x4= 6x1, x2, x3, x4 0 We havem= 2andn= 4 Thus, when determining the basic solutions, we set 2 indices to zero Suppose we choose indices{1,2}and setx1= 0andx2= 0 Substituting these in the equations yields.

4 X3= 24andx4= 6 Corresponding basic solution isx= (0,0,24,6)and it is a basicfeasible solution In this solution,x1andx2are nonbasic variables, whilex3andx4arebasic variablesOperations Research Methods10 Lecture 4 Determining an optimal solution by exhaustive search From the LP theory (take course IE 411), and optimal value of an LPproblem is always attained at a corner point Thus, we can find the optimal value and an optimal solution by Generating a list of all basic solutions Crossing out infeasible solutions Computing the objective value for each feasible solution Choosing the basic feasible solutions with the best value (min or max)Operations Research Methods11 Lecture 4 Example: Reddy Mikks caseBasic VariablesBasic SolutionFeasibility StatusObjective Valuex1, x23, , x36, 12 InfeasibleXXXx1, x44,0 Feasible20x2, x33,12 Feasible15x2, x46, 6 InfeasibleXXXx3, x424,6 Feasible0 Thus, the optimal solution isx1= 3,x2= ,x3= 0, andx4= 0and the optimal value isz= 21.

5 In this case, we have only one solution Map each of the basic solutions to the corner point in the plot of theReddy Mikks Constraint SetOperations Research Methods12 Lecture 4 Reddy Mikks: Objective & Optimal SolutionObjectivez= 5x1+ 4x2to be maximizedOperations Research Methods13 Lecture 4 Search Bottleneck: Large number of the basic solutions Blind search can be time consuming (inefficient) Given an LP in the Standard form withmequations andnvariables,there aren(n 1)(n 2) (n m+ 1)m!many basic solutions Saym= 4andn= 8, then there are 70 solutions It is hard to manually list them all and find the best We will use a more efficient method (simplex method) to perform a smarter search (selectively moving to a better point)Operations Research Methods14